Question

How may values of $\theta ϵ[0,2\pi ]$ satisfies the equation $2cos\theta +sec\theta =5tan\theta$.
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Answer 1

We have a sec θ in the expression. We will multiply by $cos\theta$ to remove $sec\theta$. [When we have $sin\theta$ and $cosec\theta$ or $tan\theta$ and $cot\theta$, we will try to remove $cosec\theta$ and $cot\theta$]
Multiplying by cose throughout, 2 $co{s}^2\theta +1=5sin\theta$
Now, if we express $co{s}^2\theta =1-si{n}^2\theta$, we will get a quadratic in $sin\theta$. We can solve the quadratic to find our solutions.

$2(1-si{n}^2\theta )+1=5sin\theta$
$\Rightarrow 2si{n}^2\theta +5sin\theta -3=0$
$2si{n}^2\theta -sin\theta +6sin\theta -3=0$
$2sin\theta (sin\theta -\frac{1}{2})+6(sin\theta -\frac{1}{2})=0$
$\Rightarrow sin\theta =\frac{1}{2}orsin\theta =-3$
$Butsin\theta ca{n}^{\prime }tbe-3$
$\Rightarrow sin\theta =\frac{1}{2}$
There will be two values of $\theta$ for which $sin\theta =\frac{1}{2}$ between 0 and $2\pi$
$i.e.\theta =\frac{\pi }{6}and\theta =\frac{5\pi }{6}$